Problem: What is the extraneous solution to these equations? $\dfrac{x^2 - 1}{x - 10} = \dfrac{x + 11}{x - 10}$
Solution: Multiply both sides by $x - 10$ $ \dfrac{x^2 - 1}{x - 10} (x - 10) = \dfrac{x + 11}{x - 10} (x - 10)$ $ x^2 - 1 = x + 11$ Subtract $x + 11$ from both sides: $ x^2 - 1 - (x + 11) = x + 11 - (x + 11)$ $ x^2 - 1 - x - 11 = 0$ $ x^2 - 12 - x = 0$ Factor the expression: $ (x + 3)(x - 4) = 0$ Therefore $x = -3$ or $x = 4$ The original expression is defined at $x = -3$ and $x = 4$, so there are no extraneous solutions.